If a snack cake contains 450 food calories and Carla

A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of

zaharov [31]

Answer:

a) pH = 2.88

b) pH = 4.598

c) pH = 5.503

d) pH = 8.788

e) pH = 12.097

Explanation:

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.75 E-5 = [H3O+]*[CH3COO-] / [CH3COOH]

a) 0.0 mL KOH:

mass balance:

⇒ C CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M

charge balance:

⇒ [H3O+] = [CH3COO-]

⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])

⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0

⇒ [H3O+] = 1.314 E.3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.88

b) 5.0 mL KOH:

CH3COOH + KOH ↔ CH3COONa + H2O

∴ C CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)

⇒ C CH3COOH = 0.05 M

∴ C KOH = ((5 E-3)(0,200))/(0.025+5 E-3) = 0.033 M

mass balance:

⇒ C CH3COOH + C KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M

charge balance:

⇒ [H3O+] + [K+] = [CH3COO-]

⇒ [CH3COO-] = [H3O+] + 0.033

⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))

⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])

⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]

⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0

⇒ [H3O+] = 2.523 E-5 M

⇒ pH = 4.598

equivalent point:

(C*V)acid = (C*V)base

⇒ (0.100 M)*(0.025 L) = (0.200 M)( Vbase)

⇒ Vbase = 0.0125L = 12.5 mL

c) 10.0 mL KOH:

∴ C CH3COOH = 0.0143 M

∴ C KOH = 0.057 M

as in the previous point, starting from the mass and charge balances, we obtain:

⇒ [H3O+] = 3.1386 E-6 M

⇒ pH = 5.503

d) 12.5 mL KOH:

at the equivalence point, there is complete salt formation, then the pH is calculated through the salt:

CH3COO- + H2O ↔ CH3COOH - OH-

∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]

∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M

mass balance:

⇒ 0.066 = [CH3COOH] + [CH3COO-]..........(1)

charge balance:

⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M.........(2)

∴ [K+] = C CH3COO- = 0.066 M

(1) = (2):

⇒ [OH-] = [CH3COOH].......(3)

⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])

⇒ [OH-]² + 5.714 E-10[OH-] - 3.7712 E-11 = 0

⇒ [OH-] = 6.1408 e-6 m

⇒ pOH = 5.212

⇒ pH = 14 - pOH = 8.788

d) 15.0 mL KOH:

after the equivalence point there is salt and excess base (OH-); ph is calculated from excess base:

⇒ C KOH = ((0.015)(0.200) - (0.025)(0.100)) / (0.025 + 0.015) = 0.0125 M

⇒ [OH-] ≅ C KOH = 0.0125 M

⇒ pOH = 1.903

⇒ pH = 12.097

8 0

3 years ago

(a) Calculate the mass of CaCl2·6H2O needed to prepare 0.125 m CaCl2(aq) by using 500. g of water. (b) What mass of NiSO4·6H2O m

Alinara [238K]

Answer:

(a) 13.7 g.

(b) 28.91 g.

Explanation:

molality (m) is the no. of moles of solute dissolved in 1.0 kg of solvent.

∴ m = (no. of moles of solute)/(mass of water (kg))

∴ m = (mass/molar mass of solute)/(mass of water (kg)).

(a) Calculate the mass of CaCl₂·6H₂O needed to prepare 0.125 m CaCl₂(aq) by using 500. g of water.

∵ m = (mass/molar mass of CaCl₂·6H₂O)/(mass of water (kg)).

m = 0.125 m, molar mass of CaCl₂·6H₂O = 219.0757 g/mol, mass of water = 500.0 g = 0.5 kg.

∴ 0.125 m = (mass of CaCl₂·6H₂O / 219.0757 g/mol)/(0.5 kg).

∴ mass of CaCl₂·6H₂O = (0.125 m)(219.0757 g/mol)(0.5 kg) = 13.7 g.

(b) What mass of NiSO₄·6H₂O must be dissolved in 500. g of water to produce 0.22 m NiSO₄(aq)?

∵ m = (mass/molar mass of NiSO₄·6H₂O)/(mass of water (kg)).

m = 0.22 m, molar mass of NiSO₄·6H₂O = 262.84 g/mol, mass of water = 500.0 g = 0.5 kg.

∴ 0.125 m = (mass of NiSO₄·6H₂O / 262.84 g/mol)/(0.5 kg).

∴ mass of NiSO₄·6H₂O = (0.22 m)(262.84 g/mol)(0.5 kg) = 28.91 g.

6 0

4 years ago

Enter the atomic symbol, including mass number and atomic number, for iodine-125.

ICE Princess25 [194]

Answer:

Explanation:

Iodine - 125

The atomic symbol of iodine is ¹²⁵₅₃ I

The symbol for iodine is I

The atomic number of iodine is 53,

and the atomic mass of iodine is 125 .

The representation of the atomic symbol is as, the atomic mass is written in uppercase and the atomic number is written in lower case , followed by the symbol of the element .

Iodine is a radio active element , used for many biological process .

It is the second largest -lived radioisotope of iodine .

The first is iodine-129 .

3 0

3 years ago

How many polar bonds are found in this molecule

Law Incorporation [45]

A bond is non polar if it is between same atoms and polar if it is between different atoms.

Same atoms are like two dogs of same strength pulling a bone towards towards each other. But when it’s different atoms it’s like a big dog and small dog then the bone is more towards bigger dog. So it’s the same way in bonds.

Bonds are made up of electrons, when the more stronger pulling atom is present than other the electrons are more towards it and as a result we have polar bond. There is development of a kind of a negative pole and a positive pole.

The stronger atom has electrons towards itself so it has a little more negative charge while the other atom has positive charge. This makes bond polar.

So just look for bond between two different atoms, it would be polar.

Look at the pic below to see the answer.

Marked with green is bond between same atoms... one carbon and another carbon so it is not polar and test marked with blue are polar.

Well the answer should have been 10 but since the bonds at 3 and 8 are two of same type we count only one of them.

The answer is 8... well the answer should be 10 otherwise... discuss it with ur teacher

6 0

3 years ago

Determine the volume of methane (ch 4 ) gas needed to react completely with 0.660 l of o 2 gas to form methanol (ch 3 oh).

Sloan [31]

Balance Chemical Equation for this reaction is,

2 CH₄ + O₂ → 2CH₃OH

According to this eq, 22.4 L (1 moles) of Oxygen requires 44.8 L (2 mole) CH₄ for complete reaction.
So, the volume of CH₄ required to consume 0.66 L of O₂ is calculated as,

22.4 L O₂ required to consume = 44.8 L CH₄
0.660 L O₂ will require = X L of CH₄

Solving for X,
X = (44.8 L × 0.660 L) ÷ 22.4 L

X = 1.320 L of CH₄

Result:
1.320 L of CH₄ gas is needed to react completely with 0.660 L of O₂ gas to form methanol (CH₃OH).

4 0

3 years ago